Acidified `KMnO_4` oxidises `H_2O_2` to `H_2o` and `O_2` . The coefficient of `H_2O_2` in the balanced chemical reaction of `KMnO_4` with `H_2O_2` in the presence of dil `H_2SO_4` is

Explanation

To balance the reaction between acidified KMnO4 and H2O2, we need to follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced chemical equation for the reaction is:
\[ \text{KMnO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{H}_2\text{O} + \text{O}_2 \]

### Step 2: Identify the oxidation states
In this reaction, we need to identify the oxidation states of the elements involved:
- In KMnO4, Mn is in the +7 oxidation state.
- In H2O2, the oxidation state of O is -1.
- In H2O, O is in the -2 oxidation state.
- In O2, O is in the 0 oxidation state.

### Step 3: Determine the changes in oxidation states
- Mn changes from +7 in KMnO4 to +2 in MnSO4.
- O in H2O2 changes from -1 to 0 in O2 and -2 in H2O.

### Step 4: Write the half-reactions
1. Reduction half-reaction (for Mn):
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

2. Oxidation half-reaction (for H2O2):
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^- \]

### Step 5: Balance the electrons
To balance the electrons transferred in both half-reactions, we need to multiply the oxidation half-reaction by 5:
\[ 5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10e^- \]

### Step 6: Combine the half-reactions
Now, we add the balanced half-reactions:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- + 5\text{H}_2\text{O}_2 \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{O}_2 + 10\text{H}^+ + 10e^- \]

### Step 7: Simplify the equation
After canceling out the electrons and simplifying, we arrive at the balanced equation:
\[ 2\text{KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O} + 5\text{O}_2 \]

### Step 8: Identify the coefficient of H2O2
From the balanced equation, we see that the coefficient of H2O2 is **5**.

### Final Answer
The coefficient of H2O2 in the balanced chemical reaction is **5**.

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